Răspuns:
Explicație pas cu pas:
[tex]\sin x+\cos x=2\sqrt{2}\sin x\cos x |()^2\\\sin^2x+2\sin x\cos x+\cos^2x=8\sin^2x\cos^2 x\\1+\sin2x=2\sin^22x\\\sin 2x\stackrel{not}{=}t,t\in{[-1,1]}\\1+t=2t^2\\2t^2-t-1=0\\\Delta=9\Rightarrow\sqrt{\Delta}=3\\t_1=\dfrac{1+3}{4}=1\\t_2=\dfrac{1-3}{4}=\dfrac{-2}{4}=-\dfrac{1}{2}\\\texttt{Mai departe inlocuiesti t cu }\sin2x\texttt{ si obtii solutiile.}[/tex]
